Distortion of a signal consisting of two periodic components

Question

We have a signal $$x(t)=U_1\cos\omega_1t+U_2\cos\omega_2t$$ whose frequencies are $f_1=100\,\text{Hz}$ and $f_2=600\,\text{Hz}$. This signal is being discretisated in time using unipolar rectangular pulse with 1 V amplitude, length $\tau$ and period $T=\frac{1}{f_0}$. Can we reconstruct this signal without distortion if we use ideal low-pass filter if sampling frequency is:

a) $f_0=1\,\text{kHz}$
b) $f_0=1.3\,\text{kHz}$

My attempt:

when we want to tell whether we can or cannot reconstruct original signal when sampling with specific frequency we compare sampling frequency with the frequency of the given signal, sampling frequency must be at least as twice big as signal frequency (NYQUIST criterion). However i cannot determine frequency of this signal since it has two components.

Since we are doing time discretisation with rectangular pulse we want to find fourier transform of $f(t)=x(t)s(t)$ where $s(t)=\sum_{n=-\infty}^{\infty} \frac{\tau}{T}sinc(n\omega_0\tau/2)e^{jn\omega_0t}$.

I know that multiplication in time is convolution in frequency but i don't know how to implement it in this case since $X(j\omega)=\frac{U_1}{2}2\pi\delta(\omega -\omega_1)+\frac{U_1}{2}2\pi\delta(\omega +\omega_1) + \frac{U_2}{2}2\pi\delta(\omega -\omega_2) + \frac{U_2}{2}2\pi\delta(\omega +\omega_2)$

How can i do that and how can i tell, when i find fourier transform, if signal can be reconstructed or not? Any help appreciated!

Answer 1

The question is trying to make you understand Aliasing

The low-pass filter is introduced (as a hint) to imply that the output is band-limited to half the Sampling frequency (Why ? See below for aliases).
This would allow you to understand how out of band signals can sneak into your sampled signals if left unchecked. Look at p5 here

You should read in detail about sampling. Skipping the real math, as rule of thumb you can do as below to answer your questions quick

a) for $f_0=1\,\text{kHz}$ the signal cannot be reconstructed

Once you sample your signal with a rate you create aliases at frequencies
$\pm f_{signal}\pm n*f_{sampling}$
where n can take any positive value

If you calculate all combinations for all your frequencies you end up with
[100 400 600 900 1100 1600 1900 2100 2900 ...]

Since you'd sampled with 1000Hz, and are using an (ideal) LPF at 500Hz, you'd take all components from the set above between [0-500Hz] in which case you end up with [100 400].

Do you see the 400Hz odd man out here?
This is termed as the alias of the 600Hz signal.

So, for an input of [100 600] and you end up with [100 400].

b) for $f_0=1.3\,\text{kHz}$ reconstruction works

Repeat steps above.

You'll find that the only components between [0-650Hz] are [100 600].

There you go !